Serway solucionario

Chapter 24 Solutions
24.1 (a) (b) (c) ΦE = EA cos θ = (3.50 × 103)(0.350 × 0.700) cos 0° = 858 N · m2/C

θ = 90.0°

ΦE = 0

ΦE = (3.50 × 103)(0.350 × 0.700) cos 40.0° = 657 N · m2/C

24.2

ΦE = EA cos θ = (2.00 × 104 N/C)(18.0 m2)cos 10.0° = 355 kN · m2/C

24.3

ΦE = EA cos θ A = π r 2 = π (0.200)2 = 0.126 m2 5.20 × 105 = E (0.126) cos 0° E = 4.14 × 106 N/C = 4.14 MN/C

24.4The uniform field enters the shell on one side and exits on the other so the total flux is zero .

24.5

(a)

A′ = (10.0 cm)( 30.0 cm) A′ = 300 cm 2 = 0.0300 m 2 Φ E, A ′ = EA′ cos θ Φ E, A ′ = 7.80 × 10 4 (0.0300) cos 180° Φ E, A ′ = − 2.34 kN ⋅ m 2 C
3 0.0 cm

(

)

0.0 cm

0.0˚

(b)

Φ E, A = EA cos θ = 7.80 × 10 4 ( A) cos 60.0°  10.0 cm  2 2 A = ( 30.0 cm )( w ) = ( 30.0cm )  = 600 cm = 0.0600 m  cos 60.0°  Φ E, A = 7.80 × 10 4 (0.0600) cos 60° =

(

)

(

)

+ 2.34 kN ⋅ m 2 C

(c)

The bottom and the two triangular sides all lie parallel to E, so Φ E = 0 for each of these. Thus,

© 2000 by Harcourt, Inc. All rights reserved.

Chapter 24 Solutions Φ E, total = − 2.34 kN ⋅ m 2 C + 2.34 kN ⋅ m 2 C + 0 + 0 + 0 = 0

33

© 2000 by Harcourt,Inc. All rights reserved.

34

Chapter 24 Solutions

24.6

(a) (b) (c)

Φ E = E ⋅ A = (ai + b j) ⋅ A i = aA Φ E = (ai + bj) ⋅ Aj = bA Φ E = (ai + bj) ⋅ Ak = 0

24.7

Only the charge inside radius R contributes to the total flux. Φ E = q / e0

24.8

Φ E = EA cos θ through the base Φ E = ( 52.0)( 36.0) cos 180° = –1.87 kN · m2/C Note the same number of electric field lines gothrough the base as go through the pyramid's surface (not counting the base). For the slanting surfaces, Φ E = +1.87 kN ⋅ m 2 / C

24.9

The flux entering the closed surface equals the flux exiting the surface. The flux entering the left side of the cone is Φ E = ∫ E ⋅ dA = E R h . This is the same as the flux that exits the right side of the cone. Note that for a uniform field only the crosssectional area matters, not shape.

*24.10 (a)

E=

k eQ r2 (8.99 × 109)Q , (0.750)2 But Q is negative since E points inward.

8.90 × 102 =

Q = – 5.56 × 10–8 C = – 55.6 nC (b) The negative charge has a spherically symmetric charge distribution.

24.11

(a) (b)

ΦE =

qin ( +5.00 µ C − 9.00 µ C + 27.0 µ C − 84.0 µ C) = = – 6.89 × 106 N · m2/C = – 6.89 MN · m2/C e0 8.85 × 10 −12 C 2 /N ⋅ m 2

Since the net electric flux is negative, more lines enter than leave the surface.

Chapter 24 Solutions qin e0 ΦE = −2Q + Q = e0 − Q e0

35

24.12

ΦE =

Through S1

Through S2 Φ E =

+ Q−Q = 0 e0 ΦE = −2Q + Q − Q 2Q = − e0 e0

Through S3

Through S4

ΦE = 0

24.13

(a)

One-half of the total flux created by the charge q goes through the plane. Thus, Φ E,plane = q 1 1 q  Φ E, total =   = 2e0 2 2  e0 

(b)

The square looks like an infinite plane to a charge very close to the surface. Hence, Φ E, square ≈ Φ E, plane = q 2e0

(c)

The plane and the square look the same to the charge.

24.14

The flux through the curved surface is equal to the flux through the flat circle, E0 π r 2 .

24.15

(a)

+Q 2 e0 –Q 2 e0

Simply considerhalf of a closed sphere.

(b)

(from ΦΕ, total = ΦΕ, dome + ΦΕ, flat = 0)

© 2000 by Harcourt, Inc. All rights reserved.

36

Chapter 24 Solutions

Goal Solution A point charge Q is located just above the center of the flat face of a hemisphere of radius R, as shown i n Figure P24.15. What is the electric flux (a) through the curved surface and (b) through the flat face? G: FromGauss’s law, the flux through a sphere with a point charge in it should be Q e0 , so we should expect the electric flux through a hemisphere to be half this value: Φ curved = Q 2e0 . Since the flat section appears like an infinite plane to a point just above its surface so that half of all the field lines from the point charge are intercepted by the flat surface, the flux through this section should...